Mongolian Mathematical Olympiad

Let $\angle BAM = \angle CAN$ and $\angle MAN = x$. By the property of bisector $\frac{BL}{LC} = \frac{AB}{AC}$. If the area of $\triangle ABM$ is $S_{ABM}$ and that of $\triangle ANC$ is $S_{ANC}$ then $S_{ABM} = \frac{1}{2}AB \cdot AM \sin \varphi$, $S_{ANC} = \frac{1}{2}AC \cdot AN \sin \varphi$. From this
$$\frac{S_{ABM}}{S_{ANC}} = \frac{AB \cdot AM}{AC \cdot AN} = \frac{BM}{NC}$$
Also
$$\frac{S_{ABN}}{S_{AMC}} = \frac{\frac{1}{2}AB \cdot AN \sin(\varphi + x)}{\frac{1}{2}AM \cdot AC \sin(\varphi + x)} = \frac{AB \cdot AN}{AM \cdot AC} = \frac{BN}{MC}$$
It follows from the two equalities that $\frac{BM \cdot BN}{NC \cdot MC} = \frac{AB^2}{AC^2}$. By AM-GM inequality,
$$\frac{BM}{MC} + \frac{NB}{NC} \geq 2\sqrt{\frac{BM \cdot NB}{MC \cdot NC}} = 2\sqrt{\frac{AB^2}{AC^2}} = 2 \cdot \frac{AB}{AC} = 2 \cdot \frac{BL}{LC}$$
Equality holds for $M \equiv N \equiv L$.